%调用newton插值法确定经过这些点的9阶多项式
X=(0:1:9);%0表示1994，9表示2003
Y=[67.052,68.008,69.803,72.024,73.400,72.063,74.669,74.487,74.065,76.777];
x=linspace(0,20);
M=1;
[y,R,A,C,L] = newton(X, Y, x, M)
%y是在向量x处的插值，误差限R，n次newton多项式L以及系数常量C
plot(X, Y, 'or', x, y, '.k');  
legend('样本点','牛顿插值估算');
x=18 %预测2010年全世界每天的石油产量
f =-(1695361414010883*x^9)/2305843009213693952 + (4227396721652201*x^8)/144115188075855872 - (8819278816003891*x^7)/18014398509481984 + (4994493159566287*x^6)/1125899906842624 - (3333749145844049*x^5)/140737488355328 + (5324582419842471*x^4)/70368744177664 - (2460605999175011*x^3)/17592186044416 + (4766208684428403*x^2)/35184372088832 - (7120293268944661*x)/140737488355328 + 16763/250
 